Question

Calculate the lattice energy of CaCl₂ from the given data

\[\ce{Ca_{(s)} + Cl2_{(g)} -> CaCl2_{(s)}}\] ∆`"H"_"f"^0` = − 795 kJ mol⁻¹

Sublimation: \[\ce{Ca_{(s)} -> Ca-{(g)}}\] ∆`"H"_1^0` = + 121 kJ mol⁻¹

Ionisation: \[\ce{Ca_{(g)} -> Ca^2+_{(g)} + 2e^-}\] ∆`"H"_2^0` = + 2422 kJ mol⁻¹

Dissociation: \[\ce{Cl2_{(g)} -> 2Cl_{(g)}}\] ∆`"H"_3^0` = + 242.8 kJ mol⁻¹

Electron affinity: \[\ce{Cl_{(g)} + e^- -> Cl^-_{(g)}}\] ∆`"H"_4^0` = −355 kJ mol⁻¹

Answer 1

∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + 2∆H₄ + u

− 795 = 121 + 2422 + 242.8 + (2 × − 355) + u

− 795 = 2785.8 – 710 + u

− 795 = 2075.8 + u

u = − 795 – 2075.8

u = −2870.8 KJ mol⁻¹